Differentiation Question 417
Question: If $ y={{\sin }^{-1}}\frac{\sqrt{(1+x)}+\sqrt{(1-x)}}{2} $ , then $ \frac{dy}{dx}= $
Options:
A) $ \frac{1}{\sqrt{(1-x^{2})}} $
B) $ -\frac{1}{\sqrt{(1-x^{2})}} $
C) $ -\frac{1}{2\sqrt{(1-x^{2})}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ x=\cos \theta $
$ 1+\cos \theta =2{{\cos }^{2}}( \frac{\theta }{2} ),\ \ 1-\cos \theta =2{{\sin }^{2}}( \frac{\theta }{2} ) $
$ \therefore y={{\sin }^{-1}}[ \frac{1}{\sqrt{2}}{ \cos ( \frac{\theta }{2} )+\sin ( \frac{\theta }{2} ) } ] $
$ ={{\sin }^{-1}}\sin ( \frac{\theta }{2}+\frac{\pi }{4} )=\frac{\theta }{2}+\frac{\pi }{4} $
Therefore $ y=\frac{1}{2}{{\cos }^{-1}}x+\frac{\pi }{4}\Rightarrow \frac{dy}{dx}=\frac{-1}{2\sqrt{1-x^{2}}} $ .
BETA
